What are the solutions to the equation 2x^2+16x+28=0 ?

The solutions to the equation 2x^2+16x+28=0 are x=-4+sqrt(2),x=-4-sqrt(2)

Find the zeros of 2x^2+16x+28=0

The zeros of 2x^2+16x+28=0 are x=-4+sqrt(2),x=-4-sqrt(2)

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2 x^{2} + 16 x + 28 = 0

x = - 4 + 2, x = - 4 - 2

Solution steps

2 x^{2} + 16 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 28}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 \cdot 28 = 42

x_{1, 2} = \frac{- 16 \pm 4 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 4 2}{2 \cdot 2}, x_{2} = \frac{- 16 - 4 2}{2 \cdot 2}

x = \frac{- 16 + 4 2}{2 \cdot 2} : - 4 + 2

x = \frac{- 16 - 4 2}{2 \cdot 2} : - 4 - 2

The solutions to the quadratic equation are:

x = - 4 + 2, x = - 4 - 2

x, x^{2} y + x y^{2} - 2 x + 2 y - 2 = 0

3 x^{2} + 8 x = - 4

y^{2} + 36 = 0

2 x^{2} + 7 x + 6 = 0

9 x^{2} - 25 x - 30 = 0

What are the solutions to the equation 2x^2+16x+28=0 ?
The solutions to the equation 2x^2+16x+28=0 are x=-4+sqrt(2),x=-4-sqrt(2)
Find the zeros of 2x^2+16x+28=0
The zeros of 2x^2+16x+28=0 are x=-4+sqrt(2),x=-4-sqrt(2)